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Cynosure-EPR Group Title

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  • 2 years ago
  • 2 years ago

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  1. phi Group Title
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    let b= (1,1,1) let n be a scalar we can say b= nL + e where e is a vector orthogonal to L |dw:1328148406952:dw|

    • 2 years ago
  2. phi Group Title
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    use L^T(b - nL)= 0 (L^T means L transpose. e and L perpendicular) to find the scalar n n = L^T b/(L^T L) to find the reflection of b = nL + e, go in the other direction b reflected = nL - e

    • 2 years ago
  3. phi Group Title
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    e= b - nL so b_reflected = nL - b + nL = 2nL -b

    • 2 years ago
  4. Cynosure-EPR Group Title
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    Lemme give it a go

    • 2 years ago
  5. Cynosure-EPR Group Title
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    Can you explain L^T? in attempting to find n?

    • 2 years ago
  6. phi Group Title
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    the T means transpose. L^T L is just L dot L, and L^T b is L dot b (dot product)

    • 2 years ago
  7. Cynosure-EPR Group Title
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    Okay.. so I'm extremely close. I've gotten (-1/9) of: [11] [ 1 ] [11] The book has the same, but positive. I'm guessing my math is just off somewhere.

    • 2 years ago
  8. phi Group Title
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    I get n = L dot b / (L dot L) = 5/9 and e = b - nL = (1,1,1) - (5/9) (2,1,2) = (-1, 4, -1)/9 b_reflected = nL - e = (5/9) (2,1,2) - (-1, 4, -1)/9 = (10,5,10)/9 + (1, -4, 1)/9 b_reflected = (11,1,11)/9 Here's a picture http://www.wolframalpha.com/input/?i=plot+vector+%281%2C1%2C1%29+%2C+vector+%281%2C2%2C1%29%2C++vector%2811%2F9%2C1%2F9%2C11%2F9%29%2C+vector+%28-11%2F9%2C-1%2F9%2C-11%2F9%29

    • 2 years ago
  9. Cynosure-EPR Group Title
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    Aha! You are a Godsend. <3

    • 2 years ago
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